# Z domain transfer function to difference equation

I want to convert this transfer function: $$\ \frac{2\cdot(z-0.5)\cdot(z-0.6)}{z-1}$$ to a difference equation, but I have no idea how.

Can someone help me?

Thanks,

Arjon

• Hi and welcome to dsp.stackexchange! Can you tell us, what you have tried so far and what has come up in class or lecture? Then we will be more willing to help you and you will learn more than from someone just posting the correct answer. – Max Apr 16 '19 at 13:57
• I don't follow any classes about this topic, but I need it for a project. I came up with this answer but I have no idea if it is correct: y[n-1]-y[n] = 2x[n] * (x[n-1] - 0.5x[n]) * (x[n-1] - 0.6x[n]). Can you tell me if this is correct? – Arjon Arts Apr 16 '19 at 13:59
• Possible duplicate of How do I find the difference equation from a transfer function? – MBaz Apr 16 '19 at 15:01

If $$x[n]$$ is the input of your discrete-time system and $$y[n]$$ is the output, then the transfer fucntion H(z) is written as: $$H(z)= \frac{Y(z)}{X(z)}$$ where $$X(z)=Z(x[n])\; \;\; \;,\; \;\; \;\; \;Y(z)=Z(y[n])$$ So we get: $$\frac{Y(z)}{X(z)} = \frac{2(z-0.5)(z-0.6)}{z-1}\Rightarrow (z-1)Y(z)=2(z-0.5)(z-0.6)X(z)$$ $$\Rightarrow zY(z)-Y(z)=2z^2X(z)-2,2zX(z)+0,6X(z)$$ At this point we apply Inverse Z Transform and make use of both Linearity and Time Shifting properties. So we end up with the following difference equation:$$\mathbf{y[n+1]-y[n]=2 \cdot x[n+2]-2.2 \cdot x[n+1]+0.6 \cdot x[n]}$$